\(\int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx\) [1395]

Optimal result

Integrand size = 17, antiderivative size = 100 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=-\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}-\frac {3 d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} \sqrt {b c-a d}} \]

[Out]

-1/2*(d*x+c)^(3/2)/b/(b*x+a)^2-3/4*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(1/2
)-3/4*d*(d*x+c)^(1/2)/b^2/(b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {43, 65, 214} \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=-\frac {3 d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} \sqrt {b c-a d}}-\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2} \]

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^3,x]

[Out]

(-3*d*Sqrt[c + d*x])/(4*b^2*(a + b*x)) - (c + d*x)^(3/2)/(2*b*(a + b*x)^2) - (3*d^2*ArcTanh[(Sqrt[b]*Sqrt[c +
d*x])/Sqrt[b*c - a*d]])/(4*b^(5/2)*Sqrt[b*c - a*d])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps \begin{align*} \text {integral}& = -\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac {(3 d) \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx}{4 b} \\ & = -\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac {\left (3 d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 b^2} \\ & = -\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac {(3 d) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 b^2} \\ & = -\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}-\frac {3 d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} \sqrt {b c-a d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=-\frac {\sqrt {c+d x} (2 b c+3 a d+5 b d x)}{4 b^2 (a+b x)^2}+\frac {3 d^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{4 b^{5/2} \sqrt {-b c+a d}} \]

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^3,x]

[Out]

-1/4*(Sqrt[c + d*x]*(2*b*c + 3*a*d + 5*b*d*x))/(b^2*(a + b*x)^2) + (3*d^2*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[
-(b*c) + a*d]])/(4*b^(5/2)*Sqrt[-(b*c) + a*d])

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {d^{2} \left (-\frac {\sqrt {d x +c}\, \left (5 b d x +3 a d +2 b c \right )}{d^{2} \left (b x +a \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 b^{2}}\)	\(79\)
derivativedivides	\(2 d^{2} \left (\frac {-\frac {5 \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {3 \left (a d -b c \right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(97\)
default	\(2 d^{2} \left (\frac {-\frac {5 \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {3 \left (a d -b c \right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(97\)

[In]

int((d*x+c)^(3/2)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/4*d^2/b^2*(-(d*x+c)^(1/2)*(5*b*d*x+3*a*d+2*b*c)/d^2/(b*x+a)^2+3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/(
(a*d-b*c)*b)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (80) = 160\).

Time = 0.24 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.83 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\left [\frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{4} c - a^{3} b^{3} d + {\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \, {\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}, \frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{4} c - a^{3} b^{3} d + {\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \, {\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}\right ] \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[1/8*(3*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*
b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(2*b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x
+ c))/(a^2*b^4*c - a^3*b^3*d + (b^6*c - a*b^5*d)*x^2 + 2*(a*b^5*c - a^2*b^4*d)*x), 1/4*(3*(b^2*d^2*x^2 + 2*a*b
*d^2*x + a^2*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (2*b^3*c^2 +
a*b^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + c))/(a^2*b^4*c - a^3*b^3*d + (b^6*c - a*b^5*d
)*x^2 + 2*(a*b^5*c - a^2*b^4*d)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\text {Timed out} \]

[In]

integrate((d*x+c)**(3/2)/(b*x+a)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\frac {3 \, d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, \sqrt {-b^{2} c + a b d} b^{2}} - \frac {5 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} - 3 \, \sqrt {d x + c} b c d^{2} + 3 \, \sqrt {d x + c} a d^{3}}{4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2} b^{2}} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

3/4*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) - 1/4*(5*(d*x + c)^(3/2)*b*d^2
- 3*sqrt(d*x + c)*b*c*d^2 + 3*sqrt(d*x + c)*a*d^3)/(((d*x + c)*b - b*c + a*d)^2*b^2)

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.35 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\frac {3\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{4\,b^{5/2}\,\sqrt {a\,d-b\,c}}-\frac {\frac {5\,d^2\,{\left (c+d\,x\right )}^{3/2}}{4\,b}+\frac {3\,d^2\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{4\,b^2}}{b^2\,{\left (c+d\,x\right )}^2-\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (c+d\,x\right )+a^2\,d^2+b^2\,c^2-2\,a\,b\,c\,d} \]

[In]

int((c + d*x)^(3/2)/(a + b*x)^3,x)

[Out]

(3*d^2*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(4*b^(5/2)*(a*d - b*c)^(1/2)) - ((5*d^2*(c + d*x)^(3
/2))/(4*b) + (3*d^2*(a*d - b*c)*(c + d*x)^(1/2))/(4*b^2))/(b^2*(c + d*x)^2 - (2*b^2*c - 2*a*b*d)*(c + d*x) + a
^2*d^2 + b^2*c^2 - 2*a*b*c*d)